If g(x) = 3x + √x, what is the value of g(d^2 + 6d + 9)?

[OliviaP316]

The question below is from MGMAT Guide 2 Algebra, 6th edition, page 179, question 7:

If g(x) = 3x + √x, what is the value of g(d^2 + 6d + 9)?

I get to this part:
g((d+3)^2) = 3 (d+3)^2 + √(d+3)^2

My question: why do I have to take both a positive AND negative root for √(d+3)^2—and therefore have two answers to g((d+3)^2)—when it is under a square root sign? I thought if a number was under a square root sign, you always only take the positive root.

Please explain. Thanks!

[RonPurewal]

The question below is from MGMAT Guide 2 Algebra, 6th edition, page 179, question 7:

If g(x) = 3x + √x, what is the value of g(d^2 + 6d + 9)?

I get to this part:
g((d+3)^2) = 3 (d+3)^2 + √(d+3)^2

My question: why do I have to take both a positive AND negative root for √(d+3)^2—and therefore have two answers to g((d+3)^2)—when it is under a square root sign? I thought if a number was under a square root sign, you always only take the positive root.

Please explain. Thanks!

you are correct, in that "√(stuff)" always represents a non-negative number.

BUT...
you seem to be assuming that (d + 3) is always non-negative (≥ 0), and that (–d – 3) is always non-positive (≤ 0).

these things aren't true.

• if d is any number GREATER than –3, then, (d + 3) is positive and (–d – 3) is negative.
thus, in this case, the square root works out to (d + 3).

HOWEVER,
• if d is any number LESS than –3, then, (d + 3) is NEGATIVE, and (–d – 3) is POSITIVE.
thus, in this case, the square root is actually (–d – 3).

(if d is actually –3, then the square root is just √0 = 0.)

in both of the first two cases, yes, you are taking the positive root -- but "the positive root" is a different one of those expressions in each case.

[OliviaP316]

Thanks, Ron!

[RonPurewal]

you're welcome.